convolution of functions whose integral is known












0












$begingroup$


Is there a general approach to solve the convolution
begin{align}
(f*g)(x) & = int_{-infty}^infty f(t)g(x-t),dt
end{align}
if the individual integrals $int_{-infty}^infty f(t),dt, ; int_{-infty}^infty g(t),dt $ are known?



(In my particular case $f$ and $g$ have support on $[0, T]$ only.)



Thanks!






integration convolution







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$endgroup$












  • $begingroup$
    No, the convolution is not a function of the integrals alone.
    $endgroup$
    – Giuseppe Negro
    Dec 19 '18 at 10:57
















0












$begingroup$


Is there a general approach to solve the convolution
begin{align}
(f*g)(x) & = int_{-infty}^infty f(t)g(x-t),dt
end{align}
if the individual integrals $int_{-infty}^infty f(t),dt, ; int_{-infty}^infty g(t),dt $ are known?



(In my particular case $f$ and $g$ have support on $[0, T]$ only.)



Thanks!






integration convolution







share|cite|improve this question









$endgroup$












  • $begingroup$
    No, the convolution is not a function of the integrals alone.
    $endgroup$
    – Giuseppe Negro
    Dec 19 '18 at 10:57














0












0








0





$begingroup$


Is there a general approach to solve the convolution
begin{align}
(f*g)(x) & = int_{-infty}^infty f(t)g(x-t),dt
end{align}
if the individual integrals $int_{-infty}^infty f(t),dt, ; int_{-infty}^infty g(t),dt $ are known?



(In my particular case $f$ and $g$ have support on $[0, T]$ only.)



Thanks!






integration convolution







share|cite|improve this question









$endgroup$




Is there a general approach to solve the convolution
begin{align}
(f*g)(x) & = int_{-infty}^infty f(t)g(x-t),dt
end{align}
if the individual integrals $int_{-infty}^infty f(t),dt, ; int_{-infty}^infty g(t),dt $ are known?



(In my particular case $f$ and $g$ have support on $[0, T]$ only.)



Thanks!






integration convolution




integration convolution






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asked Dec 19 '18 at 10:38









FilteruserFilteruser

32




32












  • $begingroup$
    No, the convolution is not a function of the integrals alone.
    $endgroup$
    – Giuseppe Negro
    Dec 19 '18 at 10:57


















  • $begingroup$
    No, the convolution is not a function of the integrals alone.
    $endgroup$
    – Giuseppe Negro
    Dec 19 '18 at 10:57
















$begingroup$
No, the convolution is not a function of the integrals alone.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 10:57




$begingroup$
No, the convolution is not a function of the integrals alone.
$endgroup$
– Giuseppe Negro
Dec 19 '18 at 10:57










2 Answers
2






active

oldest

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0












$begingroup$

Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.



    If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $fmathbf1_{[0,T]}$ and $g$ by $gmathbf1_{[0,T]}$.



    Now observe that: $$f(x)mathbf1_{[0,T]}(x)g(x-t)mathbf1_{[0,T]}(x-t)neq0implies$$$$ tin[0,T]cap[x-T,x]=[max(0,x-T),min(T,x)]$$
    showing that it is handsome to discern cases:




    • If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.

    • If $xin[0,T]$ then it is enough to integrate over $[0,x]$.

    • If $xin[T,2T]$ then it is enough to integrate over $[x-T,T]$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      0












      $begingroup$

      Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.






          share|cite|improve this answer









          $endgroup$



          Certainly not. The two integrals are mere constants, which do not allow to retrieve any useful information about $f$ and $g$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 11:21









          Yves DaoustYves Daoust

          129k675227




          129k675227























              0












              $begingroup$

              It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.



              If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $fmathbf1_{[0,T]}$ and $g$ by $gmathbf1_{[0,T]}$.



              Now observe that: $$f(x)mathbf1_{[0,T]}(x)g(x-t)mathbf1_{[0,T]}(x-t)neq0implies$$$$ tin[0,T]cap[x-T,x]=[max(0,x-T),min(T,x)]$$
              showing that it is handsome to discern cases:




              • If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.

              • If $xin[0,T]$ then it is enough to integrate over $[0,x]$.

              • If $xin[T,2T]$ then it is enough to integrate over $[x-T,T]$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.



                If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $fmathbf1_{[0,T]}$ and $g$ by $gmathbf1_{[0,T]}$.



                Now observe that: $$f(x)mathbf1_{[0,T]}(x)g(x-t)mathbf1_{[0,T]}(x-t)neq0implies$$$$ tin[0,T]cap[x-T,x]=[max(0,x-T),min(T,x)]$$
                showing that it is handsome to discern cases:




                • If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.

                • If $xin[0,T]$ then it is enough to integrate over $[0,x]$.

                • If $xin[T,2T]$ then it is enough to integrate over $[x-T,T]$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.



                  If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $fmathbf1_{[0,T]}$ and $g$ by $gmathbf1_{[0,T]}$.



                  Now observe that: $$f(x)mathbf1_{[0,T]}(x)g(x-t)mathbf1_{[0,T]}(x-t)neq0implies$$$$ tin[0,T]cap[x-T,x]=[max(0,x-T),min(T,x)]$$
                  showing that it is handsome to discern cases:




                  • If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.

                  • If $xin[0,T]$ then it is enough to integrate over $[0,x]$.

                  • If $xin[T,2T]$ then it is enough to integrate over $[x-T,T]$.






                  share|cite|improve this answer









                  $endgroup$



                  It is just a matter of finding the integral on RHS and knowledge of the integrals you mention is not useful.



                  If $f$ and $g$ have support $[0,T]$ then the result will be the same if $f$ is replaced by $fmathbf1_{[0,T]}$ and $g$ by $gmathbf1_{[0,T]}$.



                  Now observe that: $$f(x)mathbf1_{[0,T]}(x)g(x-t)mathbf1_{[0,T]}(x-t)neq0implies$$$$ tin[0,T]cap[x-T,x]=[max(0,x-T),min(T,x)]$$
                  showing that it is handsome to discern cases:




                  • If $x<0$ or $x>2T$ then the integrand is the zero function so that $(f*g)(x)=0$.

                  • If $xin[0,T]$ then it is enough to integrate over $[0,x]$.

                  • If $xin[T,2T]$ then it is enough to integrate over $[x-T,T]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 11:16









                  drhabdrhab

                  102k545136




                  102k545136






























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